Removing critical nodes from a graph: complexity results and polynomial algorithms for the case of bounded treewidth

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1 Removing critical nodes from a graph: complexity results and polynomial algorithms for the case of bounded treewidth B. Addis M. Di Summa A. Grosso July 6, 011 Abstract We consider the problem of deleting a limited number of nodes from a graph in order to minimize a connectivity measure between the surviving nodes. We prove that the problem is NP-complete even on quite particular types of graph, and define a dynamic programming recursion that solves the problem in polynomial time when the graph has bounded treewidth. We also extend this polynomial algorithm to several variants of the problem. Keywords: Critical node problem, treewidth, complexity, dynamic programming. 1 Introduction This paper deals with the following type of problems, which we call Critical Nodes Problems (CNPs): an undirected graph G = (V,E) with nodes V = {1,,...,n} and edges E {uv: u,v V} is given, and we aim to select a limited subset of nodes S V to be deleted in order to minimize a connectivity measure in the residual subgraph G[V \ S] (i.e., the subgraph of G induced by V \ S). We call the elements of S critical nodes, since their removal maximally impairs the connectivity of G. In this work we focus on a CNP where a cost w i is specified for deleting each node i V and a budget W > 0 for such operations is given as input. If removing S V induces a residual graph G[V \ S] whose connected components have node sets (depending on S) C 1,C,...,C p, the problem calls for determining S in order to p ( ) Ci minimize f(s) = (1) i=1 subject to w i W. () i S Dipartimento di Informatica, Università degli Studi di Torino, Corso Svizzera 185, Torino, Italy (addis@di.unito.it). Dipartimento di Matematica Pura e Applicata, Università degli Studi di Padova, Via Trieste 63, 3511 Padova, Italy (disumma@math.unipd.it). Dipartimento di Informatica, Università degli Studi di Torino, Corso Svizzera 185, Torino, Italy (grosso@di.unito.it). 1

2 The objective function (1) counts the number of node pairs that are still connected by at least one path in G after the nodes in S have been deleted. The issue of removing elements from a graph in order to impair its connectivity is present in a number of problems, in literature as well as in practical applications. The so-called flow-interdiction models, since the seminal work of Wollmer [4], deal with deleting arcs in order to minimize the maximum amount of flow that can be shipped through a network, a finite resource budget being allocated for arc deletion operations. The basic interdiction models deal with a single commodity source-sink flow on a directed capacitated network. Wood [5] also considers extensions to undirected graphs and multicommodity flows. Recently Smith and Lim [1] specifically tackle multicommodity interdiction models. Interdiction models usually call for a multi-level mixed-integer linear program to be solved; the inner max-flow problem is often handled by dualizing it. The problem of determining the maximum network fragmentation under removal of nodes appearsinthestudyof complex networks, forexampleinalbertet al. [1]. Borgatti [10]formulated a family of problems also covering the CNP for identifying so-called key players in social networks; the key players are those whose absence induces maximum fragmentation in the network. From the point of view of combinatorial optimization, problem (1) () is well formalized by Arulselvan et al. [3], in the special case where w i = 1 for all i V. These authors envisage applications in the development of immunization strategies for populations against diseases or for immunization of computer networks against malicious software or viruses. Immunizing the set of critical nodes limits the ability of the virus to spread along the paths of the underlying network. In [3] they suggest, among other applications, the CNP model as an alternative to classical immunization strategies see [11, 6] for example that usually focus on immunization of nodes with high degree, a policy that can be greedy and shortsighted. Boginski and Commander [9] apply the same model to locate the critical nodes in protein-protein interaction graphs, for applications in biology. From an engineering point of view, deleting nodes or edges from a graph in order to induce maximum fragmentation is of interest in assessing the robustness of a network structure when some form of attack is deployed on its elements. An edge-deletion problem with objective function (1) is modeled in [18] in order to investigate the robustness of a transportation infrastructure. Myung and Kim[19] propose a branch and cut approach for the same problem. With similar motivations, Dinh et al. [14] work on telecommunication networks, and tackle the problem of detecting subsets of node or arcs (they call them node-disruptors and arcdisruptors) in a directed network, in order to determine the minimum number of (ordered) pairs of nodes that are still able to establish a connection between them after deletion of such subsets. The objective function (1) is not the only (dis-)connectivity measure considered in the literature; Oost et al. [0] consider the problem of deleting nodes in order to minimize the size of the largest residual connected component max{ C 1,..., C p }. The number p of residual connected components (to be maximized) is a suitable measure as well. Arulselvan et al. [4] again also consider applications in telecommunications where a network has to be fragmented in connected components of limited size, through node deletions. Throughout the paper we refer to formulation (1) () as CNP, unless otherwise stated. TheCNPis knowntobenp-hardongeneral graphs[3]andpolynomiallysolvableontrees via dynamic programming [13]. A greedy procedure is proposed in the work by Borgatti [10]; a slightly more sophisticated, but quite effective heuristic is developed by Arulselvan et al. [3]

3 for the case of unit deletion costs. We note that, as far as objective (1) is considered, the CNP could be transformed into a multicommodity network interdiction problem. The same is not true (or, at least, not straightforward) for the other connectivity measures cited above. In our analysis, we avoid network flows and the linear programming framework of the interdiction problems, while focusing on the combinatorial structure of the CNP itself. This paper offers complexity results and exact algorithms for the CNP. We establish the NP-hardness of the CNP even on graphs with very special structure namely the cases of split graphs, bipartite graphs and complements of bipartite graphs and give inapproximability results on general graphs and split and bipartite graphs as well. We then provide a dynamic programming recursion that solves the CNP when a tree decomposition of the graph G = (V,E) is available. We also show that the same dynamic programming scheme can be adapted to handle the different objective functions mentioned above, and certain edge-deletion (instead of node-deletion) problems as well. A tree decomposition (first introduced by Robertson and Seymour [, 3]) is a mapping of a graph into a tree whose vertices X 1,X,...,X N (also called bags) are subsets of V; such a decomposition of the graph is generally not unique. A graph is said to have treewidth κ if it admits a tree decomposition where X 1, X,..., X N κ + 1. The concept of tree decomposition has proven to be a powerful tool in dynamic programming (see for example [6]) and many combinatorial optimization problems that are N P-hard on general graphs can be solved in polynomial time through dynamic programming when the graph treewidth is bounded by a constant not depending on the size of the graph. We prove that this is indeed the case for the CNP, and that polynomial-time solvability on graphs with bounded treewidth is maintained also for the problem extensions discussed above (edge deletion, different objective functions). This generalizes and extends the results given in [13] for the tree case. Graphs with bounded treewidth also include, among others, the trees, all series-parallel graphs, all outerplanar graphs and Halin graphs (a far longer list is given in [7]). The paper is structured as follows. In Section we give NP-hardness results for the CNP. In Section 3 we deal with the difficulty of approximating the optimal solution of the CNP in polynomial time. In Section 4 we recall the basics of tree decomposition and introduce the key concept of connected component configuration, which is instrumental for the dynamic programming algorithm defined in Section 5. In Section 6 we extend the dynamic programming recursion to CNPs with different connectivity measures, and in Section 7 we sketch how edge deletion problems are reducible to CNPs, and hence can be solved with an analogous dynamic programming procedure. Conclusions are drawn in Section 8. Complexity results Throughout this section we consider the CNP in the special case where w i = 1 for all i V, hence the budget constraint () amounts to requiring S K, with a given K = W V. Establishing N P-hardness for this case obviously handles the complexity of the more general formulation (1) (). 3

4 It can be easily argued that the CNP generalizes the well-known vertex cover problem on a graph G = (V,E): indeed a subset S V with S K satisfies f(s) = 0 if and only if S is a vertex cover of G of cardinality at most K. This immediately establishes the following known result. Proposition 1 The CNP is NP-hard on general graphs. A somewhat more complicated proof appears in [3]. It is well-known that the vertex cover problem can be solved in polynomial time on some special classes of graphs. The most important case is probably that of bipartite graphs. Here we prove that, on the contrary, the CNP is NP-hard even on bipartite graphs. We establish N P-hardness also on other classes of graphs (split graphs and complements of bipartite graphs), for which the vertex cover is trivial..1 Split graphs Asplit graphis agraphg = (V 1,V ;E) whosevertex set V can bepartitioned intotwo subsets V 1,V such that V 1 induces a clique and V is a stable set. Establishing the NP-completeness of the CNP on split graphs is an instrumental result for handling the bipartite case. Given S V, the induced subgraph G[S] of a split graph G has at most one connected component containing more than one node. We call this component the nontrivial connected component of G[S] (if such a component exists). Then the CNP on a split graph amounts to finding a subset S V of given cardinality such that the nontrivial connected component of G[S] is as small as possible (or all surviving nodes are isolated). Lemma Let G = (V 1,V ;E) be a split graph and consider the CNP on G where the number K of nodes to be removed satisfies 0 K V 1. Then there is an optimal solution such that only nodes in V 1 are removed from G. Furthermore, given any optimal solution, an equivalent solution satisfying this condition can be constructed in polynomial time. Proof. Given an optimal solution to the CNP, let S the set of nodes that are removed from G. Supposethat S V and let v S V. Since S is an optimal solution, the residual graph G[V \S] must contain at least one neighbor of v (otherwise it would be more convenient to keep v in the graph and remove any node belonging to the nontrivial connected component of the residual graph). Let w be a neighbor of v that belongs to the residual graph. Then, if we keep v in the graph and remove w instead, we find a solution S which is at least as good as the original one. Thus, since the original solution was optimal, this solution has to be optimal as well. Note that S V = S V 1. By iterating this procedure we eventually find an optimal solution S such that S V =. Proposition 3 The CNP is NP-hard even on split graphs. Proof. We show that if there exists a polynomial-time algorithm that solves the CNP on split graphs, then there is also a polynomial-time algorithm for the maximum edge biclique problem (MEBP), which is known to be NP-hard [1]. The MEBP is the following problem: given a bipartite graph G = (V 1,V ;E), find a biclique (i.e., a complete bipartite subgraph) in G with the maximum number of edges. It is clear that the MEBP can be solved in polynomial time if for each k = 1,..., V 1 the 4

5 following subproblem P k can be solved in polynomial time: find a biclique B in G satisfying V(B) V 1 = k, such that the number of edges in E(B) is maximized. (In order to always have a feasible solution to P k, we allow a biclique B to have no nodes in one of the two sides of the bipartition in this case the number of edges of B is obviously equal to zero.) Note that in P k the condition that the number of edges in E(B) is maximized can be replaced with the condition that the number of nodes in V(B) is maximized. Next we show that P k is an instance of the CNP on a split graph. Fix k {1,..., V 1 } and define S k = {S V 1 : S = k}. We formulate problem P k by defining a function f k as follows: for S S k, f k (S) is the maximum number of nodes in a biclique B of G such that V(B) V 1 = S. Note that solving problem P k is equivalent to finding a set S S k maximizing f k. Let Ḡ be the bipartite complement of G (i.e., Ḡ = (V,Ē) with Ē = {v 1v : v 1 V 1,v V,v 1 v / E}), and let G be the split graph obtained from Ḡ by placing an edge between every pair of nodes in V 1. Let I be the instance of the CNP on G where the number of nodes to be removed is V 1 k. By Lemma, we can restrict our attention to those solutions of I in which only nodes of V 1 are removed from G. Furthermore, we can assume wlog that exactly V 1 k nodes of V 1 are removed from G in an optimal solution. In other words, there is an optimal solution of instance I in which the set of nodes kept in the graph is S V for some S S k. Now, for S S k, define f (S) as the number of connected pairs of nodes in G [S V ]. The above discussion shows that solving I is equivalent to finding a set S S k minimizing f. For any nonnegative integer α k and for any S S k, we have f k (S) = α in G there are exactly α k nodes in V that are adjacent to all nodes in S in Ḡ there are exactly V (α k) nodes in V that are adjacent to at least one node in S the nontrivial connected component of G [S V ] contains exactly V (α k)+k nodes ( ) f V +k α (S) =. (3) This shows that S maximizes f k is and only if S minimizes f, thus solving P k is equivalent to solving an instance of the CNP on a split graph.. Bipartite graphs We will consider bipartite graphs of a special form, obtained as follows. Let G = (V 1,V ;E) be a split graph, where V 1 is a clique and V is a stable set. We construct a bipartite graph G starting from G and replacing every edge ij of the clique G[V 1 ] with a chain i v ij j, where v ij is a new vertex. Note that this operation is not performed for the edges linking a node in V 1 to a node in V. We denote by V the set of ( V 1 ) nodes that have been added. Then G is a bipartite graph, where the two classes are V 1 and V V. We say that G is the bipartite subdivision of G. 1 1 Note that this is slightly different from the standard definition of bipartite subdivision of a graph, as usually all edges are subdivided. 5

6 Lemma 4 Let G be the bipartite subdivision of a split graph G = (V 1,V ;E) and consider the CNP on G, where the number of nodes to be removed is K, with 0 K V 1. Then there is an optimal solution such that only nodes in V 1 are removed from G. Furthermore, given any optimal solution, an equivalent solution satisfying this condition can be constructed in polynomial time. Proof. First we show that if at most V 1 nodes are removed from G, then all the nodes in V 1 that are still in the graph belong to the same connected component. To see this, let i,j be any two nodes in V 1. In G there exist V 1 1 internally disjoint paths connecting i and j, namely the paths i v ij j and i v ih h v hj j for all h V 1 \{i,j}. It follows that it is not possible to disconnect i and j by removing at most V 1 nodes (unless i or j is removed). Now consider the CNP on G, where the number of nodes to be removed is K, with 0 K V 1, andlets bethesetofk nodesremovedaccordingtoanoptimal solution. As shownabove, thenodesinv 1 \S belongtothesameconnectedcomponentoftheresidualgraph (and this component contains more than one node). Moreover, there is a single nontrivial connected component in the residual graph. Assume that S V and let v ij S V. If both i,j S, then the solution cannot be optimal, as it would be more convenient to keep v ij in the graph and to remove some other node. If exactly one of i,j is in S (say i S), then the solution cannot be optimal, as it would be more convenient to remove j instead of v ij. Therefore both i and j are in the residual graph. Then, since i,j belong to the same connected component of the residual graph, we can equivalently remove i instead of v ij. In other words, by replacing S with S \{v ij } {i} we still have an optimal solution, where the number of nodes in S V has decreased by one unit. By iterating this process, we eventually obtain an optimal solution S V 1 V. Now, sinces V 1 V and thereis asingle nontrivial connected component in theresidual graph, in order to show that there is an equivalent solution such that only nodes in V 1 are removed from G, one can replicate the argument used in the proof of Lemma. Proposition 5 The CNP is NP-hard even on bipartite graphs. Proof. We show that the CNP on split graphs (which is NP-hard by Proposition 3) polynomially reduces to the CNP on bipartite graphs. More specifically, we prove that if G is a split graph and G is the bipartite subdivision of G, then an optimal solution to the CNP on G would immediately give an optimal solution to the CNP on G (where the number of nodes to be removed is the same). Let G = (V 1,V ;E) be a split graph and let G be its bipartite subdivision (with vertex set V(G ) = V 1 V V ). Consider the CNP on G where the number of nodes to be removed is K V 1. Let S V 1 be a subset with S = K and let α be the number of nodes in the nontrivial connected component of G[V \S]. Then the number of nodes in the nontrivial connected component of G [V(G )\S] isα+ V ( K ) = α+c, wherecis aconstant depending only on the instance. Since, by Lemma (resp., Lemma 4) there is an optimal solution to the CNP on G (resp., G ) such that only nodes in V 1 are removed, we see that an optimal solution to the latter problem gives an optimal solution to the former problem whenever K V 1. To conclude, observe that if K V 1 then the solution of the CNP on G is trivial (remove at least all the nodes in V 1 ), and if K = V 1 1 an optimal solution to the CNP on G can be found by testing the V 1 subsets of V 1 that contain exactly K nodes. 6

7 .3 Complements of bipartite graphs The complement of a bipartite graph is a graph G = (V 1,V ;E) such that each of V 1,V induces a clique, while there are arbitrary connections between the nodes in V 1 and those in V. Given such a graph G, we define G as the bipartite graph obtained by removing all the edges between nodes in V 1 and all the edges between nodes in V. In other words, G = (V 1,V ;E ), where E = {v 1 v E : v 1 V 1,v V }. Lemma 6 Let G = (V 1,V ;E) be the complement of a bipartite graph and consider the CNP on G where the number of nodes to be removed is K V 1 + V. Two cases are possible: (i) if G does not admit a vertex cover with at most K nodes, then removing any subset S of nodes with S = K gives an optimal solution to the CNP on G; (ii) if G admits a vertex cover with at most K nodes, then an optimal solution to the CNP on G is obtained by removing any subset S of nodes such that S is a vertex cover of G with S = K and V1 \S V \S (4) is as small as possible. Furthermore, all optimal solutions are of this form. Proof. If G does not admit a vertex cover with at most K nodes, then it is not possible to create two distinct connected components in G by removing only K nodes, as some edge connecting a node in V 1 to a node in V will always survive. If G admits a vertex cover with at most K nodes, then it is possible (and indeed convenient) to create two distinct connected components V 1 \S and V \S by removing the nodes in a vertex cover S with exactly K nodes (note that if K V 1 + V, exactly K nodes will be removed in any optimal solution). Since the total number of nodes in the two connected components is fixed, the best choice is to make them as balanced as possible (see [3]). Proposition 7 The CNP is NP-hard even on the complements of bipartite graphs. Proof. We show that if the CNP on the complements of bipartite graphs can be solved in polynomial time, then the constrained vertex cover problem on bipartite graphs (CVCB), which is known to be NP-complete [17], can also be solved in polynomial time. The CVCB is the following problem: given a bipartite graph H = (V 1,V ;E) and two nonnegative integers k 1 V 1, k V, determine whether there is a vertex cover of H containing at most k 1 nodes from V 1 and at most k nodes from V. Note that we can rephrase the problem by saying exactly instead of at most. By adding a suitable number of isolated nodes to either V 1 or V, one can always assume that in the CVCB the value of V 1 V is equal to some given number (provided that this number is polynomial in the size of the input). For our purpose, it will be convenient to assume that V 1 V = k 1 k, i.e., V 1 k 1 = V k. Furthermore, we can assume that k 1 +k V 1 + V, as otherwise the existence of a vertex cover with at most k 1 nodes from V 1 and k nodes from V can be easily checked in polynomial time. Given an instance of the CVCB as above, let G be the unique graph such that G = H, where G is the complement of some bipartite graph. We consider the CNP on G where the number of nodes to be removed is K := k 1 +k V 1 + V. Let S be the set of nodes removed according to an optimal solution. By the previous lemma, S is a vertex cover of 7

8 G = H if and only if H admits a vertex cover with at most K nodes. Thus, if S is not a vertex cover of H then the answer to the CVCB is negative. NowassumethatS isavertexcoverofh. Then,againbythepreviouslemma, S isavertex cover of H with K nodes minimizing expression (4). We claim that the answer to the CVCB is positive if and only if the value of (4) is zero. To see this, recall that V 1 k 1 = V k and S = k 1 + k, thus (4) is equal to zero if and only if S contains exactly k 1 nodes from V 1 and k nodes from V. Therefore, there is a vertex cover of H containing exactly k 1 nodes from V 1 and k nodes from V if and only if every optimal solution S to the CNP on G is such that expression (4) is zero. This shows that the CVCB on H can be solved by solving the CNP on G. 3 Approximability: negative results As observed at the beginning of Section, the optimal value of the CNP is zero if and only if there is a vertex cover with at most K nodes. Thus it is NP-complete to decide whether the optimal value of the CNP is zero. This immediately implies that it is NP-hard to approximate an optimal solution of the CNP within any factor, even if the factor is allowed tobeavalueγ(i) 1dependingonthespecificinstance I of thecnp(providedthatγ(i) can be computed in polynomial time). In other words, unless P = NP, there is no polynomialtime approximation algorithm that returns a solution such that APX(I) γ(i) OPT(I) for all instances I (where APX(I) and OPT(I) denote the approximate and optimal value respectively). With little more effort, one can prove that the same result holds even if an asymptotic approximation algorithm is accepted. Proposition 8 Unless P = N P, there is no polynomial-time approximation algorithm that returns a solution to the CNP such that APX(I) γ(i) OPT(I)+δ(I) (5) for all instances I, where γ(i) 1 is a function computable in polynomial time and δ(i) 0 is polynomial in the size of I. Proof. For easiness of notation, we will drop every dependence on I. Assume that an algorithm satisfying (5) exists, with δ being an integer wlog. For a graph G = (V,E) with V = n δ, consider the problem of finding the maximum stable set in G. For an integer δ k n, let I be the instance of the CNP on G where the number of nodes to be removed is n k. We run the approximation algorithm on instance I and denote by T the set of nodes that are kept in the graph according to the approximate solution (thus T = k). Two cases are possible. 1. Suppose first that APX δ. In this case T induces a subgraph of G with k nodes, in which at most δ nodes have degree greater than zero (the upper bound δ is achieved when the edges in G[T] form a matching of cardinality δ). Then the nodes in G[T] that have degree zero form a stable set of G with at least k δ nodes.. Now suppose that APX > δ. In this case (5) gives OPT APX δ γ that G does not contain a stable set of size k. > 0, which means 8

9 Summarizing, for any δ k n, we can solve the following problem: either find a stable set in G whose size is at least k δ, or prove that G does not contain any stable set of size k. By running the algorithm a polynomial number of times, one can find the maximum number k such that a stable set S with k δ S k is returned. In particular, no stable set of size k +1 exists in G. If we denote by S a maximum stable set in G, we then have S S k ( k δ) = δ. This means that we could solve the stable set problem in polynomial time within polynomial absolute error. However, it is well-known that this is not possible, unless P = NP. The above proof immediately implies the following. Corollary 9 Let F be a family of graphs over which it is NP-hard to approximate the maximum stable set problem within polynomial absolute error. Then, unless P = NP, there is no polynomial-time asymptotic approximation algorithm for the CNP, even if we restrict the set of instances to F. Together with [15], the above result implies that there is no polynomial-time asymptotic approximation algorithm for the CNP even in the special cases of planar graphs, graphs with bounded degree, and even cubic planar graphs (unless P = NP). We pointed out in Section that the CNP is NP-hard even on bipartite graphs. It is not clear whether in this case a polynomial-time approximation algorithm can be found. However it is quite easy to see that a polynomial-time algorithm that solves the problem on bipartite graphs within constant absolute error (i.e., APX(I) OPT(I) δ for some constant δ 0) does not exist, unless P = NP. This is not surprising, as there are very few N P-hard problems that can be solved within constant absolute error; however, for the sake of completeness, we give the short proof of this result below. Proposition 10 It is NP-hard to solve the CNP on bipartite (resp., split) graphs within constant absolute error. Proof. Let us first consider the case of split graphs. Assume that there is a polynomial-time algorithm such that, for any instance I of the CNP on a split graph, APX(I) OPT(I) δ, where δ 0 is a constant. We show that then an exact optimal solution to the CNP on a split graph could be found in polynomial time. Firstweprovethatthealgorithmoutputsanoptimalsolutionwhenever V 1 δ+1andthe number of nodes to be removed from the graph is K V 1 (δ+1). When K V 1 (δ+1), at least δ + 1 nodes of V 1 are kept in the graph. Then, since every induced subgraph of a split graph has at most one nontrivial connected component, we have OPT(I) = ( α ) and APX(I) = ( β ) for some δ + 1 α β. If the solution returned by the algorithm is not optimal, then β α+1 and APX(I) OPT(I) = ( ) β ( ) α ( ) α+1 ( ) α = α δ +1, a contradiction. Thus the solution returned by the algorithm is optimal whenever V 1 δ+1 and k V 1 (δ +1). When the above condition does not hold, i.e., when V 1 δ or K V 1 δ, it is sufficient to test all the subsets of V 1 containing exactly K nodes (the number of these subsets is polynomial, as δ is a constant). 9

10 To prove the result for bipartite graphs, one has to adapt the above arguments to the case in which the graph is the bipartite subdivision of a split graph. 4 Tree decompositions, treewidth and connected component configurations In this section we introduce the key concepts that will be used in Section 5 to develop our dynamic programming algorithm. 4.1 Tree decompositions We first recall a few definitions and basic properties concerning tree decompositions and the treewidth of a graph. Given a graph G = (V,E), a tree decomposition of G is a tree T whosevertices X 1,...,X N (also called bags) are subsets of V, satisfying the following properties: (a) N i=1 X i = V; (b) for each edge uv E, there exists a bag X i containing both u and v; (c) for each node u V, the subgraph of T induced by the bags {X i : u X i } is connected. We will always assume that the tree decomposition T is rooted at X 1 this allows us to establish a parent-descendants relation between nodes. Note that every graph admits a tree decomposition (just take a single bag containing all the vertices of V). The width of the tree decomposition T is defined as max i X i 1. The treewidth of G is the minimum width of a tree decomposition of G. Finding the treewidth of a general graph is an NP-hard problem []. However, for any given constant κ, there is a linear time algorithm that checks whether the treewidth of a graph G is at most κ, and (if this is the case) constructs a tree decomposition of G of width at most κ [8]. Given a bag X i of T, we will use notation V i to indicate the set of nodes (of G) obtained by merging all the bags being descendants of X i (including X i itself). We will use the following easy and well-known result: Lemma 11 Let X i be a bag of T and let X j,x k be two children of X i. Then for any u V j \X i and v V k \X i, we have u v and uv / E. Proof. If u = v then by (c) u would belong to X i, a contradiction. If uv E then combining (b) and (c) we can argue that either u or v (or both) belongs to X i, a contradiction. A tree decomposition T of G is called a nice tree decomposition of G if it satisfies the following additional constraints: (d) every bag of T has at most two children; (e) if X i has two children X j,x k, then X i = X j = X k ; (f) if X i has exactly one child X j, then either X j = X i \{v} for somev X i, or X j = X i {v} for some v V \X i ; 10

11 G C 1 C S = {1,,3,5,6,7} A 1 = C 1 S = {1,,3,5} A = C S = {6,7} α = {(A 1,1),(A,)} Figure 1: Illustration of the definition of CCC: α is the CCC of S with respect to the graph G, which has two connected components C 1,C. (g) every leaf-bag of T has cardinality one. Given atreedecompositionof Gof widthτ, it isalways possibletoconstructinpolynomial time a tree decomposition of G of width τ that satisfies (d) (f), where the number of bags is O(n) (see, e.g., [16]). If condition (g) is also required, then the number of bags is O(τn). 4. Connected component configurations In order to develop our algorithm, we need to introduce the concept of Connected Component Configuration (CCC in the following). Let G = (V,E) be a graph and take S V. We define the CCC of S with respect to G as the following unordered sequence of ordered pairs: α = {( C 1 S, C 1 \S ),..., ( C p S, C p \S )}, where C 1,...,C p are the connected components of G that have nonempty intersection with S. In other words, for every C l, α indicates which nodes of S and how many nodes of V \S belong to C l (see Figure 1 for an example). Given a finite set S and a nonnegative integer r, we denote by Γ(S,r) the set of all objects α of the form α = {(A 1,a 1 ),...,(A p,a p )}, where (a) A 1,...,A p are nonempty subsets that form a partition of S; (b) a 1,...,a p are nonnegative integers such that a 1 + +a p r. The elements of Γ(S,r) will be called r-potential CCCs of S. Note that (a) (b) are necessary conditions for the existence of a graph G = (V,E), with S V and V \S r, such that α is the CCC of S with respect to G. Any such graph G will be called a realization of α. We will make use of three operations involving potential CCCs, which are introduced and discussed in the next subsections Restriction of a potential CCC Let α Γ(S,r) be an r-potential CCC of a nonempty finite set S and take any v S. Assume that α = {(A 1,a 1 ),...,(A p,a p )}, where without loss of generality v A 1. We define the restriction of α to S \{v} as the following r-potential CCC of S \{v}: α v = {(A 1 \{v},a 1 +1),(A,a ),...,(A p,a p )}, (6) 11

12 S = {,4,6,9,7} α = {( {,4,6,9},3),({7},1) } S \{9} = {,4,6,7} α 9 = {( {,4,6},4),({7},1) } Figure : Restriction of the CCC α of S to S \{9}. where the first pair (A 1 \{v},a 1 +1) should be omitted if A 1 = {v} (as in this case A 1 \{v} = ). Lemma 1 Given a nonempty finite set S and α Γ(S,r), if graph G is a realization of α then G is also a realization of α v for all v S. Proof. If v is removed from S but kept in G, the connected components of G are unchanged. Modifying (A 1,a 1 ) into (A 1 \{v},a 1 +1) reflects the fact that v has been moved from S to V \S. An illustration of Lemma 1 is given in Figure. 4.. Extension of a potential CCC We define an extension operation for an r-potential CCC that intuitively relates to the augmentation of a graph G by adding a new node (and possibly connecting it to other nodes). Let α Γ(S,r) bean r-potential CCC of a finite set S, whereα = {(A 1,a 1 ),...,(A p,a p )}. Consider a new element v / S together with a subset Q S. We define the extension of α to v with neighborhood Q to be the following r-potential CCC of S {v}: where L = {l : A l Q }. ext(α,v,q) = {(A l,a l ) : l / L} {( {v} l L A l, l L a l)}, (7) Lemma 13 Given a finite set S and α Γ(S,r), let G = (V,E) be a realization of α. Then any graph G obtained by adding a new node v to G and connecting it arbitrarily to a set of other nodes N(v) S is a realization of ext(α,v,n(v)). Proof. Set Q = N(v) and define L as above: L = {l : A l Q }. Then l L if and only if A l contains a neighbor of v in G. Since v merges all those connected components C l (respectively, sets A l of α) that intersect N(v), while it does not affect the C l s (resp. A l s) that do not intersect N(v), the CCC of S {v} with respect to G is exactly ext(α,v,q). An illustration of Lemma 13 is given in Figure 3. 1

13 G G S = {,4,6,7} α = { ({},),({4,6},1),({7},1) } N(9) = {,4} ext(α,9,{,4}) = {( {,4,6,9},3),({7},1) } Figure 3: Extension of the CCC α to node 9 with neighborhood Q = {,4} Sum of two potential CCCs In order to develop a dynamic programming algorithm for the CNP, we need to define an operation that allows to combine two potential CCCs β 1,β into a new potential CCC α in such a way that α is uniquely determined by the information in β 1,β. We call such α the sum of β 1,β denote it by α = β 1 +β. The sum is defined as follows. 1. Assume that β t = {(B t 1,bt 1 ),...,(Bt p t,b t p t )} for t = 1,.. Build an auxiliary graph H with vertex set S and edge set defined as follows: given u,v S, H contains edge uv if and only if u,v belong to the same set Bk t for some k {1,...,p t } and some t {1,}. 3. Let A 1,...,A p be the connected components of H. 4. Define a l = b 1 k + b k for l = 1,...,p. (8) k:b 1 k A l k:b k A l 5. Define α = {(A 1,a 1 ),...,(A p,a p )}. When dealing with realizations of CCCs, the following problem arises. Consider a graph G = (V,E) and a subset S V, let U 1,U be two subsets of V containing S such that U 1 U = V. Let β 1 (resp., β ) be the CCC of S with respect to G[U 1 ] (resp., G[U ]). Let α be the CCC of S with respect to G. The example in Figure 4 shows that the knowledge of β 1,β is generally not sufficient to determine α, even if U 1 U = S. However, the following result identifies conditions under which we can safely compute α as β 1 +β. Lemma 14 Let β 1 Γ(S,r 1 ),β Γ(S,r ), where S is a finite set and r 1,r are nonnegative integers. Suppose that there exist a graph G = (V,E) such that S V and two subsets U 1,U V such that G[U 1 ] (resp., G[U ]) is a realization of β 1 (resp., β ). If U 1 U = V, U 1 U = S and there is no edge linking a node in U 1 \S with a node in U \S, then α = β 1 +β is the CCC of S with respect to G. 13

14 G G[U 1 ] G [U 1 ] S = U 1 U = {1,,3} α = { ({1,},3),({3},3) } U 1 = {1,,3,4,5,8} β 1 = { ({1,},),({3},1) } G G[U ] G [U ] S = U 1 U = {1,,3} α = { ({1,,3},6) } U = {1,,3,6,7,9} β = { ({1},0),({},1),({3},) } Figure 4: Case where it is not possible to sum two CCCs: here β 1 (resp. β ) is the CCC of S with respect to G[U 1 ] (resp. G[U ]), while α (resp. α ) is the CCC of S with respect to G (resp. G ). Because of the choice of U 1,U the same induced subgraphs G[U 1 ], G[U ] with identical β 1,β are generated from two different graphs G,G. As a result, we cannot reliably compute α or α by using only the information in β 1,β. This situation is ruled out under the hypothesis of Lemma

15 Proof. In the following we show that α as defined in steps 1 5 above is the CCC of S with respect to G. First we prove that if u,v A l for some l, then u,v belong to the same connected component of G. If u,v A l, then there exists a path P in H with u and v as endpoints. Let xy be any edge of P. Since xy is an edge of H, x and y belong to the same connected component of either G[U 1 ] or G[U ]. In both cases G contains a path connecting x and y. Since this holds for all the edges of P, we can argue that u and v are connected by a path in G, i.e., u and v belong to the same connected component of G. We now show that if u,v are nodes in S that belong to the same connected component of G, then u,v A l for some l. Let P be a path in G having u and v as endpoints. Without loss of generality, we assume that the node of P that is a neighbor of u belongs to U 1. Let P be a maximal subpath of P starting at u with the property that all the nodes of P belong to U 1, and let w be the last node of P. Note that P contains at least one edge. Also note that w S: if not, then w v (as v S) and by the maximality of P, the successor of w in P (z, say) would belong to U \ S; however this is not possible, as wz would be an edge linking a node in U 1 \S to a node in U \S, a contradiction to our assumptions. Thus w S. Furthermore, since P is contained in G[U 1 ], u and w belong to the same connected component of G[U 1 ], i.e., H contain edge uw. Now we consider the path P \P (which starts at w and ends at v) and define P as the maximal subpath of P \P starting at w with the property that all the nodes of P belong to U. Note that P contains at least one edge. By iterating this process (alternating maximal subpaths in G[U 1 ] and G[U ]), we eventually reach the final node v. This proves that there is a path in H connecting u and v, i.e., u,v A l for some l. The above shows that the family of sets {A 1,...,A p } is precisely the family of sets {C 1 S,...,C p S}, where C 1,...,C p are the connected components of G that have nonempty intersection with S. In order to conclude that α is the CCC of S with respect to G, we have to prove that a l = C l \S for l = 1,...,p. However, by construction we see that each A l is the union of some sets Bk t. Then, recalling that U 1\S and U \S share no nodes, the number of nodes in V \S that are connected to A l is given by (8). 5 A dynamic programming algorithm We are now ready to describe a dynamic programming algorithm that solves the nodeweighted version of the CNP (1) (). The algorithm works correctly on any graph G, if a nice tree decomposition of G is known. However, the running time is polynomial only if the width of the tree decomposition is bounded by a constant. We first recall the node-weighted CNP. Given a graph G = (V,E) with weights w i on the vertices, and given a budget W, the problem is to remove a subset S V of total weight w(s) W such that the numberof connected pairs of nodes in G[V \S] is as small as possible. In order to simplify notations and calculations, from now on we consider the problem in its complementary form: we want to select S V, with w(s) W, such that the number of connected pairs in G[S] is as small as possible, where W = w(v) W. Let T be a nice tree decomposition of G. Let X 1,...,X N V denote the bags of T. Recall that V i is the union of X i with all its descendant bags. We now define the recursive function that is used in our dynamic programming algorithm. Given i = 1,...,N, a subset S X i, a potential CCC α Γ(S, V i \ X i ) and an integer 15

16 0 m ( V i ), we define the following function: f i (S,α,m) = max { w(r) : S R V i \(X i \S), G[R] is a realization of α, the number of connected pairs in G[R] is m }, with f i (S,α,m) = when there is no subset R satisfying the above conditions. We denote by P i (S,α,m) the optimization problem (9). The values of f i are calculated starting from the leaves of T and proceeding in postorder. In order to have a clean presentation of the algorithm, we first state all recursive formulas and postpone the proof of their correctness to Proposition 15. However, rather than reading first all the algorithm and then the proofs of all formulas, the reader might prefer to look at the proof of the correctness of each formula right after having read the formula itself. (9) Initial conditions Let X i = {v} be a leaf-bag. We set w v if S = {v}, α = {({v},0)}, m = 0 f i (S,α,m) = 0 if S =, α =, m = 0 in all other cases. (10) Case 1 recursion Let X i be a bag with two children X i1,x i. Recall that X i = X i1 = X i. We compute f i (S,α,m) = max { f i1 (S,β 1,m 1 )+f i (S,β,m ) w(s) : β t Γ(S, V it \X i ), 0 m t ( V it ) for t = 1,, β 1 +β = α, Φ(β 1,β,m 1,m ) = m }, where, assuming that β t = {(B1 t,bt 1 ),...,(Bt p t,b t p t )} for t = 1,, and α = β 1 + β = {(A 1,a 1 ),...,(A p,a p )}, function Φ is defined as follows: p 1 ( B 1 Φ(β 1,β,m 1,m ) = m 1 +m l +b 1 ) p ( l B l +b ) p ( ) l Al +a l +. (1) l=1 Note that we do not have α as an argument of Φ: in fact Φ does not really depend on α, as α = β 1 +β can be computed through the algorithm given in Section Case recursion Let X i be a bag with a single child X j. We have to consider the two cases X j = X i \{v} for some v X i and X j = X i {v} for some v V \X i. Case.1 Assume X j = X i \{v} for some v X i and let N(v) denote the set of neighbors of v in G[V i ]. We compute f j (S,α,m) if v / S, w v +max { f j (S,α,m ) : f i (S,α,m) = α Γ(S, V j \X j ), 0 m ( V j ) (13), if v S, ext(α,v,n(v) S) = α, Ψ(α,m } ) = m We remark that f i(s,α,m) is finite only if 0 m ( V i \(X i ) \S), thus we could restrict m to satisfy this condition. However, in order to keep notation simpler, we allow m to take values from 0 up to ( ) V i. 16 l=1 l=1 (11)

17 wheres = S\{v} and, assumingthat α = {(A 1,a 1 ),...,(A p,a p )} andα = ext(α,v,n(v) S) = {(A 1,a 1 ),...,(A p,a p )}, Ψ(α,m ) is defined as follows: p ( A Ψ(α,m ) = m l +a ) l + l=1 p ( Al +a l Similarly to Case 1, Ψ does not really depend on α, as α = ext(α,v,n(v) S). l=1 ). (14) Case. If X j = X i {v} for some v V \X i, we compute { f j (S,α,m) f i (S,α,m) = max max { f j (S,α,m) : α Γ(S, V j \X j ), α v = α }, (15) where S = S {v}. Optimal value Recall that the tree decomposition T is rooted at X 1. The optimal value is given by min { m : f 1 (S,α,m) W, S X 1, α Γ(S, V \X 1 ), 0 m ( V )}. (16) As usual in dynamic programming, an optimal solution can be recovered by backtracking. Proposition 15 The dynamic program described by formulas (10) (16) solves the nodeweighted CNP, once a nice tree decomposition of the graph is given. Proof. We prove the correctness of all the formulas given above. Initial conditions. If X i = {v} is a leaf-bag, then V i = X i. It follows that the only possible feasible solution to problem P i (S,α,m) is R = S. When S = {v}, the solution R = S is feasible if and only if α = {({v},0)} (as G[R] must be a realization of α) and m = 0 (no connected pairs); when S =, the solution R = S is feasible if and only if α = and m = 0. Case 1 recursion. We now prove the correctness of formula (11). In order to show that the left-hand side of (11) is at most as large as the right-hand side, we prove that if f i (S,α,m) is finite then there exist β 1,β,m 1,m as in (11) such that f i (S,α,m) f i1 (S,β 1,m 1 )+f i (S,β,m ) w(s). Afterwards, in order to show that the lefthand side of (11) is at least as large as the right-hand side, we prove that if f i1 (S,β 1,m 1 ) and f i (S,β,m ) are finite for some β 1,β,m 1,m as in (11), then f i (S,α,m) f i1 (S,β 1,m 1 )+ f i (S,β,m ) w(s). Assume that f i (S,α,m) is finite. Then there exists R such that S R V i \(X i \S), α is the CCC of S with respect to G[R], and the number of connected pairs in G[R] is m. For t = 1,, define R t = R V it, let β t be the CCC of S with respect to G[R t ], and let m t be the number of connected pairs in G[R t ]. We have to show that (i) β 1 + β = α, (ii) Φ(β 1,β,m 1,m ) = m and (iii) f i (S,α,m) f i1 (S,β 1,m 1 )+f i (S,β,m ) w(s). (i) Since R 1 \S V i1 \X i1 = V i1 \X i and R \S V i \X i = V i \X i, by Lemma 11 we have that R 1 \S and R \S are disjoint subsets of nodes, and there is no edge linking a node in the former set to a node in the latter set. Then, since R 1 R = S, we can apply Lemma 14 with U 1 = R 1 and U = R, obtaining that β 1 +β is the CCC of S with respect to G[R 1 R ] = G[R], i.e., α = β 1 +β. 17

18 (ii)toseethatthenumberofconnectedpairsing[r](i.e., m)isequaltoφ(β 1,β,m 1,m ), the crucial observation is that every connected component of G[R 1 ] or G[R ] which does not intersect S is also a connected component of G[R] that does not intersect S, and viceversa. Let us analyze the right-hand side of (1). The first two terms (i.e., m 1 +m ) give the number of connected pairs in G[R 1 ] plus the number of connected pairs in G[R ]. With the two subsequent summations we are subtracting all the connected pairs that belong to a component of G[R 1 ] or G[R ] that intersects S. Thus we are so far counting exactly the connected pairs belonging to components of G[R 1 ] or G[R ] that are disjoint from S, i.e., the number of connected pairs belonging to a component of G[R] that is disjoint from S. We now have to add the number of connected pairs belonging to a component of G[R] that intersects S, and this number is given by the last summation. (iii) By construction of β 1,β,m 1,m, we have that R 1 (resp., R ) is a feasible solution to problem P i1 (S,β 1,m 1 ) (resp., P i (S,β,m )). Therefore, since R 1 R = R and R 1 R = S, we have f i (S,α,m) = w(r) = w(r 1 )+w(r ) w(s) f i1 (S,β 1,m 1 )+f i (S,β,m ) w(s). The above arguments show that the left-hand side of (11) is at most as large as the righthand side. To prove the other inequality, assume that both f i1 (S,β 1,m 1 ) and f i (S,β,m ) are finite for some β 1,β,m 1,m as in (11). Then, for t = 1,, there exists R t such that S R t V it \ (X i \ S), β t is the CCC of S with respect to G[R t ], and m t is the number of connected pairs in G[R t ]. Define R = R 1 R and α = β 1 + β. Then, by Lemma 14, α is the CCC of S with respect to R. Also, by setting m = Φ(β 1,β,m 1,m ), m is the number of connected pairs in G[R]. Then R is a feasible solution to problem P i (S,α,m) and f i (S,α,m) w(r) = w(r 1 )+w(r ) w(s) = f i1 (S,β 1,m 1 )+f i (S,β,m ) w(s). Case.1. We now prove the correctness of formula (13). When v / S, thecorrectness of (13) follows fromthefact that thesubproblemsp i (S,α,m) and P j (S,α,m) have precisely the same feasible solutions. Thus in the following we assume that v S. We first prove that the left-hand side of (13) is at most as large as the right-hand side. Assume that f i (S,α,m) is finite. Then there exists R such that S R V i \(X i \S), α is the CCC of S with respect to G[R], and the number of connected pairs in G[R] is m. Define R = R \ {v}, let α be the CCC of S with respect to G[R ], and let m be the number of connected pairs in G[R ]. We have to show that (i) α = ext(α,v,n(v) S), (ii) Ψ(α,m ) = m and (iii) f i (S,α,m) w v +f j (S,α,m ). (i) Refer to the definition of nice tree decomposition. By property (c), recalling that X j = X i \{v}, we have v / V j. Further, using also property (b), for the set of neighbors N(v) of v in G[V i ] we have N(v) X i. Moreover, since S R V i \(X i \S), the set of neighbors of v in G[R] is contained in S. Then, by Lemma 13, ext(α,v,n(v) S) is the CCC of S with respect to G[R], i.e., α = ext(α,v,n(v) S). (ii) To see that the number of connected pairs in G[R] (i.e., m) is equal to Ψ(α,m ), use again the facts that N(v) X i and X j = X i \ {v}. With this in mind, the basic idea is a counting argument as in Case 1. (iii) By construction of α and m, we have that R is a feasible solution to problem P j (S,α,m ). Then f i (S,α,m) = w(r) = w v +w(r ) w v +f j (S,α,m ). To prove that the left-hand side of (13) is at least as large as the right-hand side, assume that f j (S,α,m ) is finite for some α and m as in (13). Then there exists R such that S R V j \(X j \S ), α is the CCC of S with respect to G[R ], and m is the number of connected pairs in G[R ]. Define R = R {v}, α = ext(α,v,n(v) S) and m = Ψ(α,m ). 18

19 Since, by Lemma 13, α is the CCC of S with respect to R, and since m is the number of connected pairs in G[R], R is a feasible solution to problem P i (S,α,m). Hence f i (S,α,m) w(r) = w v +w(r ) = w v +f j (S,α,m ). Case.. We now prove the correctness of formula (15). We first prove that the left-hand side of (15) is at most as large as the right-hand side. Assume that f i (S,α,m) is finite. Then there exists R such that S R V i \(X i \S), α is the CCCofS withrespecttog[r], andthenumberofconnected pairsing[r]ism. Ifv / R, then R is a feasible solution to problem P j (S,α,m) and thus f i (S,α,m) = w(r) f j (S,α,m). Now assume that v R and let α be the CCC of S with respect to G[R]. By Lemma 1, α v is the CCC of S with respect to G[R], i.e., α v = α. Since R is a feasible solution to problem P j (S,α,m), we have f i (S,α,m) = w(r) f j (S,α,m). Thus in all cases the left-hand side of equation (15) is not larger than the right-hand side. Now assume that the right-hand side of (15) is finite. If f j (S,α,m) is finite, then there exists R such that S R V j \ (X j \ S ), α is the CCC of S with respect to G[R], and m is the number of connected pairs in G[R]. In this case R is clearly a feasible solution to problem P i (S,α,m), thus f i (S,α,m) w(r) = f j (S,α,m). Now assume that f j (S,α,m) is finite for some α as in (15). Then there exists R such that S R V j \(X j \ S ), α is the CCC of S with respect to G[R ], and m is the number of connected pairs in G[R ]. Since, by Lemma 1, α v = α, R is a feasible solution to problem P i (S,α,m). Then f i (S,α,m) w(r) = f j (S,α,m). Optimal value. Recall that we are interested in a subset R V with weight w(r ) W such that G[R ] contains the minimum number of connected pairs, which we denote by m. To see that m is at least as large as the expression in (16), let S = R X 1, α be the CCC of S with respect to R and m = m. Then R is a feasible solution to problem P 1 (S,α,m), thus f 1 (S,α,m) w(r ) W. Then m is at least as large as the minimum in (16). To see that m is at most as large as the expression in (16), let S,α,m be parameters for which the minimum is attained, and let R be a corresponding optimal solution of problem P 1 (S,α,m). Then R is a set of nodes with w(r) W such that G[R] contains m connected pairs, thus m m. Wenowshowthat ifthewidthofagiventreedecompositionofgisboundedbyaconstant, then the dynamic program described above requires only a polynomial number of operations. Proposition 16 Given a tree decomposition of G whose width is bounded by a constant κ, the dynamic program described by formulas (10) (16) requires only a polynomial number of operations. Proof. As recalled in Section 4, given a tree decomposition of G one can obtain in polynomial time a nice tree decomposition of G of the same width, where the number of bags is O(κn) = O(n). The recursive function f i (S,α,m) must be computed for all bags X i of T, all S X i, all α Γ(S, V i \ X i ) and all integers 0 m ( V i ). We now bound the number of possible choices of these parameters. There are O(n) possible choices for the index i. Since, for each fixed i, bag X i contains at most κ + 1 nodes of V, there are at most κ+1 possible choices of a subset S X i. As V i n, at most ( n ) +1 values of m need to be considered. 19